For the reaction
$C_{2} H_{6} \to C_{2} H_{4} + H_{2}$
The reaction enthalpy $Δ_{r} H =$ ______________ $k J m o l^{- 1}$ .(Round off to Nearest Integer].
[Given : Bond enthalpies in $k J m o l^{- 1}$ . $: C - C : 347, C = C : 611;$ $C - H : 414, H - H : 436$ ]

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answer is 128.

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Detailed Solution

$\begin{array}{l} Δ H = \sum (B E)_{R} - \sum (B E)_{P} \\ = [C - C + 6 C - H] - [C = C + 4 C - H + H - H] \\ = [347 + 6 (414)] - [611 + 4 + (414) + 436] = 128 \end{array}$

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